Archangel Macsika
Prove that for all integer n>=3, P(n+1,3) - P(n,3) = 3P(n,2)
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\\ \text{Proof: LHS}= \\ [\text{Using}\ P(n, \mu)=\frac{n !}{(n-\mu) !}] \\ \frac{(n+1) !}{(n+1-3) !}-\frac{n !}{(n-3) !} \\ = \frac{n !(n+1)}{(n-2) !}-\frac{n !}{(n-3) !} \\ = \frac{(n+1) n !}{(n-2)(n-3) !}-\frac{n !}{(n-3) !} \\ = \frac{n !}{(n-3) !}\left[\frac{(n+1)}{(n-2)}-1\right] \\ = \frac{n !}{(n-3) !}\left[\frac{n+1-n+2}{n-2}\right] \\ =\frac{n !}{(n-3) !}\left[\frac{3}{(n-2)}\right] \\ =\frac{3 n !}{(n-2)(n-3) !}=3 \frac{n !}{(n-2) !} \\ = \quad 3 P(n, 2)= \text{RHS} \\ \text{Hence proved.}